Exercise 24, § 2

Let K be a maximal ordered field; consider an ordering on the field $K(X)$ of rational functions which makes it an ordered extension of K. Show that this ordering is determined by the set A of $x \in K$ such that $x < X$ (show that the sign of every polynomial f over K is determined by A, using Prop. 9 of VI, p. 28). Conversely, show that to every set $A \subset K$ such that $y \in A$ whenever $y \leq x$ and $x \in A$, there corresponds an ordering of $K(X)$ making it an ordered extension of K and such that A is the set of $x \in K$ such that $x < X$ (same method). When A has a greatest element, or $\complement A$ has a least element, or when $A = K$ or $A = \varnothing$, then the corresponding ordering of $K(X)$ is such that $K(X)$ is not comparable to K. On the other hand, if neither A nor $\complement A$ is empty, and there is no greatest element in A nor least element in $\complement A$, then $K(X)$ is comparable to K.